Here is the logic:
y = 'var to check for'
some_var = some_loop.each do |x|
x if x == y
break if x
end
Is there a better way to write this?
Something like
x && break if x == y
Thank you in advance!
The correct answer is to use include?. eg:
found = (array_expression).include? {|x| x == search_value}
It's possible to also use each and break out on the first matched value, but the C implementation of include? is faster than a ruby script with each.
Here is a test program, comparing the performance of invoking include? on a very large array vs. invoking each on the same array with the same argument.
#!/usr/bin/env ruby
#
require 'benchmark'
def f_include a, b
b if a.include?(b)
end
def f_each_break a, b
a.each {|x| return b if x == b }
nil
end
# gen large array of random numbers
a = (1..100000).map{|x| rand 1000000}
# now select 1000 random numbers in the set
nums = (1..1000).map{|x| a[rand a.size]}
# now, check the time for f1 vs. f2
r1 = r2 = nil
Benchmark.bm do |bm|
bm.report('incl') { r1 = nums.map {|n| f_include a,n} }
bm.report('each') { r2 = nums.map {|n| f_each_break a,n} }
end
if r1.size != r2.size || r1 != r2
puts "results differ"
puts "r1.size = #{r1.size}"
puts "r2.size = #{r2.size}"
exceptions = (0..r1.size).select {|x| x if r1[x] != r2[x]}.compact
puts "There were #{exceptions.size} exceptions"
else
puts "results ok"
end
exit
Here is the output from the test:
$ ./test-find.rb
user system total real
incl 5.150000 0.090000 5.240000 ( 7.410580)
each 7.400000 0.140000 7.540000 ( 9.269962)
results ok
Why not:
some_var = (some_loop.include? y ? y : nil)
Related
The exercise I'm working on asks "Write a method, coprime?(num_1, num_2), that accepts two numbers as args. The method should return true if the only common divisor between the two numbers is 1."
I've written a method to complete the task, first by finding all the factors then sorting them and looking for duplicates. But I'm looking for suggestions on areas I should consider to optimize it.
The code works, but it is just not clean.
def factors(num)
return (1..num).select { |n| num % n == 0}
end
def coprime?(num_1, num_2)
num_1_factors = factors(num_1)
num_2_factors = factors(num_2)
all_factors = num_1_factors + num_2_factors
new = all_factors.sort
dups = 0
new.each_index do |i|
dups += 1 if new[i] == new[i+1]
end
if dups > 1
false
else
true
end
end
p coprime?(25, 12) # => true
p coprime?(7, 11) # => true
p coprime?(30, 9) # => false
p coprime?(6, 24) # => false
You could use Euclid's algorithm to find the GCD, then check whether it's 1.
def gcd a, b
while a % b != 0
a, b = b, a % b
end
return b
end
def coprime? a, b
gcd(a, b) == 1
end
p coprime?(25, 12) # => true
p coprime?(7, 11) # => true
p coprime?(30, 9) # => false
p coprime?(6, 24) # => false```
You can just use Integer#gcd:
def coprime?(num_1, num_2)
num_1.gcd(num_2) == 1
end
You don't need to compare all the factors, just the prime ones. Ruby does come with a Prime class
require 'prime'
def prime_numbers(num_1, num_2)
Prime.each([num_1, num_2].max / 2).map(&:itself)
end
def factors(num, prime_numbers)
prime_numbers.select {|n| num % n == 0}
end
def coprime?(num_1, num_2)
prime_numbers = prime_numbers(num_1, num_2)
# & returns the intersection of 2 arrays (https://stackoverflow.com/a/5678143)
(factors(num_1, prime_numbers) & factors(num_2, prime_numbers)).length == 0
end
I am trying to write a function scramble(str1, str2) that returns true if a portion of str1 characters can be rearranged to match str2, otherwise returns false. Only lower case letters (a-z) will be used. No punctuation or digits will be included. For example:
str1 = 'rkqodlw'; str2 = 'world' should return true.
str1 = 'cedewaraaossoqqyt'; str2 = 'codewars' should return true.
str1 = 'katas'; str2 = 'steak' should return false.
This is my code:
def scramble(s1, s2)
#sorts strings into arrays
first = s1.split("").sort
second = s2.split("").sort
correctLetters = 0
for i in 0...first.length
#check for occurrences of first letter
occurrencesFirst = first.count(s1[i])
for j in 0...second.length
#scan through second string
occurrencesSecond = second.count(s2[j])
#if letter to be tested is correct and occurrences of first less than occurrences of second
#meaning word cannot be formed
if (s2[j] == s1[i]) && occurrencesFirst < occurrencesSecond
return false
elsif s2[j] == s1[i]
correctLetters += 1
elsif first.count(s1[s2[j]]) == 0
return false
end
end
end
if correctLetters == 0
return false
end
return true
end
I need help optimising this code. Please give me suggestions.
Here is one efficient and Ruby-like way of doing that.
Code
def scramble(str1, str2)
h1 = char_counts(str1)
h2 = char_counts(str2)
h2.all? { |ch, nbr| nbr <= h1[ch] }
end
def char_counts(str)
str.each_char.with_object(Hash.new(0)) { |ch, h| h[ch] += 1 }
end
Examples
scramble('abecacdeba', 'abceae')
#=> true
scramble('abecacdeba', 'abweae')
#=> false
Explanation
The three steps are as follows.
str1 = 'abecacdeba'
str2 = 'abceae'
h1 = char_counts(str1)
#=> {"a"=>3, "b"=>2, "e"=>2, "c"=>2, "d"=>1}
h2 = char_counts(str2)
#=> {"a"=>2, "b"=>1, "c"=>1, "e"=>2}
h2.all? { |ch, nbr| nbr <= h1[ch] }
#=> true
The last statement is equivalent to
2 <= 3 && 1 <= 2 && 1 <= 2 && 2 <=2
The method char_counts constructs what is sometimes called a "counting hash". To understand how char_counts works, see Hash::new, especially the explanation of the effect of providing a default value as an argument of new. In brief, if a hash is defined h = Hash.new(0), then if h does not have a key k, h[k] returns the default value, here 0 (and the hash is not changed).
Suppose, for different data,
h1 = { "a"=>2 }
h2 = { "a"=>1, "b"=>2 }
Then we would find that 1 <= 2 #=> true but 2 <= 0 #=> false, so the method would return false. The second comparison is 2 <= h1["b"]. As h1 does not have a key "b", h1["b"] returns the default value, 0.
The method char_counts is effectively a short way of writing the method expressed as follows.
def char_counts(str)
h = {}
str.each_char do |ch|
h[ch] = 0 unless h.key?(ch) # instead of Hash.new(0)
h[ch] = h[c] + 1 # instead of h[c][ += 1
end
h # no need for this if use `each_with_object`
end
See Enumerable#each_with_object, String#each_char (preferable to String.chars, as the latter produces an unneeded temporary array whereas the former returns an enumerator) and Hash#key? (or Hash#has_key?, Hash#include? or Hash#member?).
An Alternative
def scramble(str1, str2)
str2.chars.difference(str1.chars).empty?
end
class Array
def difference(other)
h = other.each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
reject { |e| h[e] > 0 && h[e] -= 1 }
end
end
I have found the method Array#difference to be so useful I proposed it be added to the Ruby Core (here). The response has been, er, underwhelming.
One way:
def scramble(s1,s2)
s2.chars.uniq.all? { |c| s1.count(c) >= s2.count(c) }
end
Another way:
def scramble(s1,s2)
pool = s1.chars.group_by(&:itself)
s2.chars.all? { |c| pool[c]&.pop }
end
Yet another:
def scramble(s1,s2)
('a'..'z').all? { |c| s1.count(c) >= s2.count(c) }
end
Since this appears to be from codewars, I submitted my first two there. Both got accepted and the first one was a bit faster. Then I was shown solutions of others and saw someone using ('a'..'z') and it's fast, so I include that here.
The codewars "performance tests" aren't shown explicitly but they're all up to about 45000 letters long. So I benchmarked these solutions as well as Cary's (yours was too slow to be included) on shuffles of the alphabet repeated to be about that long (and doing it 100 times):
user system total real
Stefan 1 0.812000 0.000000 0.812000 ( 0.811765)
Stefan 2 2.141000 0.000000 2.141000 ( 2.127585)
Other 0.125000 0.000000 0.125000 ( 0.122248)
Cary 1 2.562000 0.000000 2.562000 ( 2.575366)
Cary 2 3.094000 0.000000 3.094000 ( 3.106834)
Moral of the story? String#count is fast here. Like, ridiculously fast. Almost unbelievably fast (I actually had to run extra tests to believe it). It counts through about 1.9 billion letters per second (100 times 26 letters times 2 strings of ~45000 letters, all in 0.12 seconds). Note that the difference to my own first solution is just that I do s2.chars.uniq, and that increases the time from 0.12 seconds to 0.81 seconds. Meaning this double pass through one string takes about six times as long as the 52 passes for counting. The counting is about 150 times faster. I did expect it to be very fast, because it presumably just searches a byte in an array of bytes using C code (edit: looks like it does), but this speed still surprised me.
Code:
require 'benchmark'
def scramble_stefan1(s1,s2)
s2.chars.uniq.all? { |c| s1.count(c) >= s2.count(c) }
end
def scramble_stefan2(s1,s2)
pool = s1.chars.group_by(&:itself)
s2.chars.all? { |c| pool[c]&.pop }
end
def scramble_other(s1,s2)
('a'..'z').all? { |c| s1.count(c) >= s2.count(c) }
end
def scramble_cary1(str1, str2)
h1 = char_counts(str1)
h2 = char_counts(str2)
h2.all? { |ch, nbr| nbr <= h1[ch] }
end
def char_counts(str)
str.each_char.with_object(Hash.new(0)) { |ch, h| h[ch] += 1 }
end
def scramble_cary2(str1, str2)
str2.chars.difference(str1.chars).empty?
end
class Array
def difference(other)
h = other.each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
reject { |e| h[e] > 0 && h[e] -= 1 }
end
end
Benchmark.bmbm do |x|
n = 100
s1 = (('a'..'z').to_a * (45000 / 26)).shuffle.join
s2 = s1.chars.shuffle.join
x.report('Stefan 1') { n.times { scramble_stefan1(s1, s2) } }
x.report('Stefan 2') { n.times { scramble_stefan2(s1, s2) } }
x.report('Other') { n.times { scramble_other(s1, s2) } }
x.report('Cary 1') { n.times { scramble_cary1(s1, s2) } }
x.report('Cary 2') { n.times { scramble_cary2(s1, s2) } }
end
I want to find the max number without running the function twice
def foo(num)
num * 10
end
def bar
x = 0
for i in 0..5
if foo(i) > x
x = foo(i) # I don't want to run foo a second time
end
end
end
How about
def bar
(1..5).map{|i| foo(i)}.max
end
This will traverse 1 to 5, and max a new enumerable with foo(i) instead of i, then return the max.
If you want the value of x:
define_method(:foo) { |x| x * 10 }
(1..5).max_by { |x| foo(x) }
#=> 5
If you want the value of f(x):
(1..5).map { |x| foo(x) }.max
#=> 50
You can save the result of the function as a variable, so you can use it later without calling the function again.
Applied to your code example, it would look like this:
#...
fooOfI = foo(i)
if fooOfI > x
x = fooOfI
end
#...
Store the result of the method in a local variable.
def bar
x = 0
for i in 0..5
foo_result = foo i
if foo_result > x
x = foo_result
end
end
end
I would do some change in your code :
def foo(num)
num * 10
end
def bar
x = 0
for i in 0..5
_,x = [foo(i),x].sort #or you can write as x = [foo(i),x].max
end
x
end
p bar
# >> 50
Elegant and simple
foo = -> x { x * 10 }
(1..5).map(&foo).max
# => 50
In one iteration (no so elegant but performs better)
def foo(num); num * 10; end;
(1..5).reduce(-1.0/0) { |a, e| f = foo(e); f > a ? f : a }
# => 50
In this question, the asker requests a solution that would insert a space every x number of characters. The answers both involve using a regular expression. How might you achieve this without a regex?
Here's what I came up with, but it's a bit of a mouthful. Any more concise solutions?
string = "12345678123456781234567812345678"
new_string = string.each_char.map.with_index {|c,i| if (i+1) % 8 == 0; "#{c} "; else c; end}.join.strip
=> "12345678 12345678 12345678 12345678"
class String
def in_groups_of(n)
chars.each_slice(n).map(&:join).join(' ')
end
end
'12345678123456781234567812345678'.in_groups_of(8)
# => '12345678 12345678 12345678 12345678'
class Array
# This method is from
# The Poignant Guide to Ruby:
def /(n)
r = []
each_with_index do |x, i|
r << [] if i % n == 0
r.last << x
end
r
end
end
s = '1234567890'
n = 3
join_str = ' '
(s.split('') / n).map {|x| x.join('') }.join(join_str)
#=> "123 456 789 0"
This is slightly shorter but requires two lines:
new_string = ""
s.split(//).each_slice(8) { |a| new_string += a.join + " " }
I'm trying to learn Ruby, and am going through some of the Project Euler problems. I solved problem number two as such:
def fib(n)
return n if n < 2
vals = [0, 1]
n.times do
vals.push(vals[-1]+vals[-2])
end
return vals.last
end
i = 1
s = 0
while((v = fib(i)) < 4_000_000)
s+=v if v%2==0
i+=1
end
puts s
While that works, it seems not very ruby-ish—I couldn't come up with any good purely Ruby answer like I could with the first one ( puts (0..999).inject{ |sum, n| n%3==0||n%5==0 ? sum : sum+n }).
For a nice solution, why don't you create a Fibonacci number generator, like Prime or the Triangular example I gave here.
From this, you can use the nice Enumerable methods to handle the problem. You might want to wonder if there is any pattern to the even Fibonacci numbers too.
Edit your question to post your solution...
Note: there are more efficient ways than enumerating them, but they require more math, won't be as clear as this and would only shine if the 4 million was much higher.
As demas' has posted a solution, here's a cleaned up version:
class Fibo
class << self
include Enumerable
def each
return to_enum unless block_given?
a = 0; b = 1
loop do
a, b = b, a + b
yield a
end
end
end
end
puts Fibo.take_while { |i| i < 4000000 }.
select(&:even?).
inject(:+)
My version based on Marc-André Lafortune's answer:
class Some
#a = 1
#b = 2
class << self
include Enumerable
def each
1.upto(Float::INFINITY) do |i|
#a, #b = #b, #a + #b
yield #b
end
end
end
end
puts Some.take_while { |i| i < 4000000 }.select { |n| n%2 ==0 }
.inject(0) { |sum, item| sum + item } + 2
def fib
first, second, sum = 1,2,0
while second < 4000000
sum += second if second.even?
first, second = second, first + second
end
puts sum
end
You don't need return vals.last. You can just do vals.last, because Ruby will return the last expression (I think that's the correct term) by default.
fibs = [0,1]
begin
fibs.push(fibs[-1]+fibs[-2])
end while not fibs[-1]+fibs[-2]>4000000
puts fibs.inject{ |sum, n| n%2==0 ? sum+n : sum }
Here's what I got. I really don't see a need to wrap this in a class. You could in a larger program surely, but in a single small script I find that to just create additional instructions for the interpreter. You could select even, instead of rejecting odd but its pretty much the same thing.
fib = Enumerator.new do |y|
a = b = 1
loop do
y << a
a, b = b, a + b
end
end
puts fib.take_while{|i| i < 4000000}
.reject{|x| x.odd?}
.inject(:+)
That's my approach. I know it can be less lines of code, but maybe you can take something from it.
class Fib
def first
#p0 = 0
#p1 = 1
1
end
def next
r =
if #p1 == 1
2
else
#p0 + #p1
end
#p0 = #p1
#p1 = r
r
end
end
c = Fib.new
f = c.first
r = 0
while (f=c.next) < 4_000_000
r += f if f%2==0
end
puts r
I am new to Ruby, but here is the answer I came up with.
x=1
y=2
array = [1,2]
dar = []
begin
z = x + y
if z % 2 == 0
a = z
dar << a
end
x = y
y = z
array << z
end while z < 4000000
dar.inject {:+}
puts "#{dar.sum}"
def fib_nums(num)
array = [1, 2]
sum = 0
until array[-2] > num
array.push(array[-1] + array[-2])
end
array.each{|x| sum += x if x.even?}
sum
end